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思路
首先搞清楚二叉搜索树的机制，左儿子的值都小于当前节点，右儿子的值都大于当前节点

然后就可以快速写出程序了，用递归很好实现

1. 相等直接return
2. 小于当前节点，直接去树节点的左子树寻找
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                <h2 class="post-title">每日一题20201129*（700. 二叉搜索树中的搜索）</h2>
                <div class="post-date">2020-11-30 20:06:48</div>
                
                <div class="post-content" v-pre>
                    <h4 id="700-二叉搜索树中的搜索"><a href="https://leetcode-cn.com/problems/search-in-a-binary-search-tree/">700. 二叉搜索树中的搜索</a></h4>
<figure data-type="image" tabindex="1"><img src="https://tva1.sinaimg.cn/large/0081Kckwgy1gl7g403logj30yu0u0tc2.jpg" alt="image-20201130195559871" loading="lazy"></figure>
<h4 id="思路">思路</h4>
<pre><code>首先搞清楚二叉搜索树的机制，左儿子的值都小于当前节点，右儿子的值都大于当前节点

然后就可以快速写出程序了，用递归很好实现

1. 相等直接return
2. 小于当前节点，直接去树节点的左子树寻找
3. 大于当前节点，去树节点的右子树寻找
</code></pre>
<pre><code class="language-Python"># Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def searchBST(self, root: TreeNode, val: int) -&gt; TreeNode:
        if root is None:
            return None
        if val == root.val:
            return root
        elif val &gt; root.val:
            return self.searchBST(root.right, val)
        else:
            return self.searchBST(root.left, val)

</code></pre>
<figure data-type="image" tabindex="2"><img src="https://tva1.sinaimg.cn/large/0081Kckwgy1gl7g6e7vfij311a0kiq57.jpg" alt="image-20201130195817737" loading="lazy"></figure>
<pre><code class="language-Go">/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func searchBST(root *TreeNode, val int) *TreeNode {
    if root == nil {
        return nil
    }
    if val &lt; root.Val {
        return searchBST(root.Left, val)
    }
    if val &gt; root.Val {
        return searchBST(root.Right, val)
    }  
    return root
}

</code></pre>
<figure data-type="image" tabindex="3"><img src="https://tva1.sinaimg.cn/large/0081Kckwgy1gl7gc03g6cj310y0esdhc.jpg" alt="image-20201130200341005" loading="lazy"></figure>

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